Câu hỏi của Le Hoang Phuong

Question

dot chay 22,4 gam Fe trong binh co chua4,48 lit khi oxi o dktc

a. sau phan ung chat nao du? Tinh khoi luong chat du

b. tinh khoi luong chat ran thu dc sau phan ung

Bùi Lan Anh · Accepted Answer

a+b, $n_{Fe\left(đb\right)}=\frac{22,4}{56}=0,4\left(mol\right)$

$n_{O_{2\left(đb\right)}}=\frac{4,48}{22,4}=0,2\left(mol\right)$

$3Fe+2O_2\rightarrow Fe_3O_4$

Theo PTHH: 3mol 2mol

$\frac{n_{Fe\left(đb\right)}}{n_{Fe\left(PTHH\right)}}$ $\frac{n_{O_{2\left(đb\right)}}}{n_{O_{2\left(PTHH\right)}}}$

$\Rightarrow\frac{0,4}{3}>\frac{0,2}{2}$

⇒ O2 hết; Fe dư

Theo PTHH: $n_{Fe_3O_4}=\frac{1}{2}n_{O_2}=\frac{1}{2}.0,2=0,1\left(mol\right)$

$\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)$Câu trả lời: a+b, $n_{Fe\left(đb\right)}=\frac{22,4}{56}=0,4\left(mol\right)$